在思考方面，这可能被称为反转。你会发现世界是圆的而不是平的。“你真的不需要知道角色的索引来处理它。”作为一名C程序员，我发现这也很难接受! 你的行“let index = letters.characters.indexOf("c")!”本身就足够了。 例如，要去掉c，你可以用…(操场粘贴)

    var letters = "abcdefg"
  //let index = letters.rangeOfString("c")!.startIndex //is the same as
    let index = letters.characters.indexOf("c")!
    range = letters.characters.indexOf("c")!...letters.characters.indexOf("c")!
    letters.removeRange(range)
    letters

然而，如果你想要一个索引，你需要返回一个实际的index而不是Int值，因为Int值对于任何实际使用都需要额外的步骤。这些扩展返回一个索引，一个特定字符的计数，以及这个游乐场插件代码将演示的范围。

extension String
{
    public func firstIndexOfCharacter(aCharacter: Character) -> String.CharacterView.Index? {

        for index in self.characters.indices {
            if self[index] == aCharacter {
                return index
            }

        }
        return nil
    }

    public func returnCountOfThisCharacterInString(aCharacter: Character) -> Int? {

        var count = 0
        for letters in self.characters{

            if aCharacter == letters{

                count++
            }
        }
        return count
    }


    public func rangeToCharacterFromStart(aCharacter: Character) -> Range<Index>? {

        for index in self.characters.indices {
            if self[index] == aCharacter {
                let range = self.startIndex...index
                return range
            }

        }
        return nil
    }

}



var MyLittleString = "MyVery:important String"

var theIndex = MyLittleString.firstIndexOfCharacter(":")

var countOfColons = MyLittleString.returnCountOfThisCharacterInString(":")

var theCharacterAtIndex:Character = MyLittleString[theIndex!]

var theRange = MyLittleString.rangeToCharacterFromStart(":")
MyLittleString.removeRange(theRange!)

我玩以下的游戏

extension String {
    func allCharactes() -> [Character] {
         var result: [Character] = []
         for c in self.characters {
             result.append(c)
         }
         return 
    }
}

直到我理解提供的一个，现在它只是字符数组

和

let c = Array(str.characters)

斯威夫特5

查找子字符串的索引

let str = "abcdecd"
if let range: Range<String.Index> = str.range(of: "cd") {
    let index: Int = str.distance(from: str.startIndex, to: range.lowerBound)
    print("index: ", index) //index: 2
}
else {
    print("substring not found")
}

查找字符索引

let str = "abcdecd"
if let firstIndex = str.firstIndex(of: "c") {
    let index: Int = str.distance(from: str.startIndex, to: firstIndex)
    print("index: ", index)   //index: 2
}
else {
    print("symbol not found")
}

与Objective-C中的NSString相比，Swift中的变量类型String包含不同的函数。Sulthan提到过，

Swift String没有实现RandomAccessIndex

你能做的是向下转换你的变量类型String到NSString(这是有效的Swift)。这将给你访问NSString中的函数。

var str = "abcdefghi" as NSString
str.rangeOfString("c").locationx   // returns 2

如果您正在寻找简单的方法来获得字符或字符串的索引，请检查这个库http://www.dollarswift.org/#indexof-char-character-int

您也可以使用另一个字符串或正则表达式模式从字符串中获取indexOf

字符串{

//Fucntion to get the index of a particular string
func index(of target: String) -> Int? {
    if let range = self.range(of: target) {
        return characters.distance(from: startIndex, to: range.lowerBound)
    } else {
        return nil
    }
}
//Fucntion to get the last index of occurence of a given string
func lastIndex(of target: String) -> Int? {
    if let range = self.range(of: target, options: .backwards) {
        return characters.distance(from: startIndex, to: range.lowerBound)
    } else {
        return nil
    }
}

}